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Trig identity.. I don't understand this one at all

Mathematics
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how do i make the left side into one big fraction
\[\huge \frac{ 1 }{ sinx+1 } - \frac{ 1 }{ sinx-1 } = \frac{ 2 }{ 1-\sin^2x}\]
typed it wrong wooops

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Other answers:

get common denominators on the left side: \(\huge \frac{ 1 }{ sinx+1 }\cdot (\color {red}{\frac{ sinx-1 }{ sinx-1 }}) - \frac{ 1 }{ sinx-1 }\cdot( \color {red}{\frac{ sinx+1 }{ sinx+1 }}) = \frac{ 2 }{ 1-\sin^2x} \)
so the left side in simplified form will be: \(\huge \frac{(sinx-1)-(sinx+1)}{sin^2x-1} \) can you simplify this?
no that's where i get stuck how would i subtract the numerator
just use distributive property to get rid of the parenthesis...
@dpaInc i get this \[\huge \frac{ \sin^2x+sinx-sinx-1 }{ sinx^2-1 }\]
how did you get that? all you're doing is subtracting....: \(\huge \frac{(sinx-1)-(sinx+1)}{sin^2x-1} \) =\(\huge \frac{1\cdot (sinx-1)-1\cdot (sinx+1)}{sin^2x-1} \) =\(\huge \frac{sinx-1-sinx-1}{sin^2x-1} \) =\(\huge \frac{-2}{sin^2x-1} \) =\(\huge \frac{2}{1-sin^2x} \)
sinx-1-sinx-1 = -2 is that a trig rule or what ? :S

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